Question

Evaluate the integral: \[ \int \frac{2x}{x^2 - 5x + 4}\, dx \]

• A. \( \frac{8}{3}\ln|x-4| - \frac{2}{3}\ln|x-1| + C \)
• B. \( \frac{2}{3}\ln|x-1| - \frac{8}{3}\ln|x-4| + C \)
• C. \( 2\ln|x-1| + \ln|x-4| + C \)
• D. \( \ln|x^2-5x+4| + C \)

Hint:

When integrating rational functions, first factor the denominator and apply partial fractions. Each resulting term usually reduces to a logarithmic integral of the form \( \int \frac{1}{x-a}dx = \ln|x-a| \).

Answer 1

The Correct Option is A

Concept: For rational functions where the degree of the numerator is less than the denominator, we use partial fraction decomposition. If \[ \frac{P(x)}{(x-a)(x-b)} \] is given, it can be written as \[ \frac{A}{x-a} + \frac{B}{x-b}. \] Integrals of the form \[ \int \frac{1}{x-a}\,dx = \ln|x-a| + C \] are then applied.
Step 1: {Factor the denominator.} \[ x^2 - 5x + 4 = (x-1)(x-4) \] Thus, \[ \int \frac{2x}{x^2-5x+4}dx = \int \frac{2x}{(x-1)(x-4)}dx \]
Step 2: {Apply partial fractions.} Assume \[ \frac{2x}{(x-1)(x-4)} = \frac{A}{x-1} + \frac{B}{x-4} \] Multiplying by \((x-1)(x-4)\): \[ 2x = A(x-4) + B(x-1) \] \[ 2x = Ax -4A + Bx - B \] \[ 2x = (A+B)x + (-4A - B) \] Comparing coefficients: \[ A + B = 2 \] \[ -4A - B = 0 \] Solving, \[ B = -4A \] \[ A -4A = 2 \] \[ -3A = 2 \Rightarrow A = -\frac{2}{3} \] \[ B = \frac{8}{3} \]
Step 3: {Substitute back into the integral.} \[ \int \frac{2x}{(x-1)(x-4)}dx = \int \left(\frac{-2/3}{x-1} + \frac{8/3}{x-4}\right)dx \] \[ = -\frac{2}{3}\int \frac{1}{x-1}dx + \frac{8}{3}\int \frac{1}{x-4}dx \]
Step 4: {Integrate.} \[ = -\frac{2}{3}\ln|x-1| + \frac{8}{3}\ln|x-4| + C \] \[ = \frac{8}{3}\ln|x-4| - \frac{2}{3}\ln|x-1| + C \]